1. Hausdorff and Non-Hausdorff Spaces Mayra Ibarra MATH101

2. What is a Topological Space?

Recall:

• A topological space is an ordered pair (X, ), where X is a set and T is a collection of subsets of X such that

• • and X

• • U V whenever U, V ;

• • { U a: a I } whenever {U a: a I } . 1

3. Separation Axioms 2

T 0: If a, b X, there exists an open set O such that either a O and bO, or b O and aO.

T 1: If a, b X, there exist open sets O a , O b containing a and b respectively, such that bO aand aO b .

4. T 2: If a, b X, there exist disjoint open sets O aand O bcontaining a and b respectively. 5. 6. T 3: If A is a closed set and b is a point not in A, there exist disjoint open sets O Aand O bcontaining A and b respectively.X 7. T 4: If A and B are disjoint closed sets in X, there exist disjoint open sets O Aand O Bcontaining A and B respectively X 8. T 5: If A and B are separated sets in X, there exist disjoint open sets O Aand O Bcontaining A and B respectively 9. Hausdorff Spaces

All metric spaces are Hausdorff 3

The real number line with the usual topology is Hausdorff 4

• Suppose we have an open set, S, of real numbers

• For eachp S we can find an >0, such that the -neighborhood ofpforms an open set inside of S.

• Supposep, q S and are separated by a distance r.We can pick 1and 2forpandq , respectively, such that r/2 1> 0 and r/2 2> 0.Thuspandqwill be separated by disjoint open sets.

10. Subspaces of Hausdorff Spaces are also Hausdorff 5

If Y X and (X, d) is Hausdorff, then any two distinct points x 1 , x 2Y will also be in X. Since x 1 and x 2are in X, there are disjoint open sets O 1and O 2containing x 1 and x 2 , respectively. Thus Y contains points that can be separated by disjoint open sets and is Hausdorff.

11. Non-Hausdorff Spaces

Zariski Topology

Let X be [0, 1][2, 3]. Define the following equivalence relation on X:

• a ~ a + 2 for all 0 a 1 except for a =

• b ~ b 2for 2 b 3 except for b = 5/2

[0] [1] 5/2 1/2 12. Proof:

Notice that every point in [0, 1] is identified with a point of [2, 3] except fpr the two points and 5/2, which remain unidentified. Thus are contained in X/~. A set in X/~ is open iff it is open in X. However, we see that there are no disjoint open sets that contain and 5/2. So X is Non-Hausdorff when considered under this equivalence relation.6

13. Regular and Normal Spaces 7

Regular Space 8-A space which is both a T 1and a T 3space

Normal Space 9 -A space which is both a T 1and a T 4space 14.

Every Normal Space is Regular

Every Regular Space is Hausdorff

Not all Hausdorff Spaces are Regular

In the definitions of Normal and Regular we can replace the use of disjoint open sets, and use disjoint closures to yield the same spaces

The same is not true for the definition of Hausdorff

15. Completely Hausdorff 10

T 2 Axiom- If a and b are two points of a topological space X, there exist open sets O aand O bcontaining a and b, respectively, such that a b=

A space that satisfies this axiom is called a completely Hausdorff space.

16. 17. Compact Hausdorff Spaces

If a Hausdorff space is compact, then it is normal, and therefore also regular.

18.

Let the sequence (S n ) in a Hausdorff Space have a limit, s. Suppose that the sequence has another limit, p, such that ps. Let U and V be disjoint neighborhoods of s and p, respectively. Then if we choose a sufficiently large n, S n U however, for a sufficiently large n, S n V. This contradicts the Hausdorff property because U V . Thus the sequence cannot have more than one limit.

19. If X is Hausdorff and (S n ) is a sequence in X that converges to a point s X, and if y is an accumulation point of the set{S n|n = 1, 2, . . .}, then s = y. 20.

Suppose sy. Then there exist open sets U, VXfor s and y respectively such that UV = . Also, since (S n ) converges to s, there exists a natural number, N such that n>N implies that S nU.

Letibe such that1

Define V i = V iin case ys i, and V i = V in case y=s i and define V as the intersection of V with a finite collection of open sets, V i . So yVa system of neighborhoods around y.

21.

Let z be such that z V and yz, then zs ifor anyisince ifi> N, s i U and UV UV = ,

And ifi N, then either y=s iz, or s i W i, and W iV W i V i= .

Thus there is a neighborhood V of y, such that no point z y of the set {S n| n = 1, 2, . . .}, belongs to V.

This contradicts the fact that y is an accumulation point of the set {S n| n = 1, 2, . . .},

thus s=y.

22. Endnotes

Greever, John.Theory and Examples of Point-Set Topology.Claremont: Waybread Publications, 1990.

Steen, Lynn Arthur and J. Arthur Seebach, Jr.Counterexamples in Topology.New York: Dover Publications Inc., 1995.

Image of axiom spaces: http://jtauber.com/2005/01/separation.png

Sneddon, I. N. Ed. Andrew H. Wallace. An Introduction To Algebraic Topology.International Series of Monographs in Pure and Applied Mathematics.V.I. New York: Pergamon Press, 1957.

Baum, John D.Elements of Point Set Topology . New York: Dover Publications Inc., 1991

Sneddon, pg. 32.

Goodman, Sue E.Beginning Topology.Belmont: Brooks/Cole, 2005.

Baum, pg. 81.

Image-http://en.wikipedia.org/wiki/Regular_space

Image-http://commons.wikimedia.org/wiki/Image:Normal_space.svg

Steen, pg. 13.

Steen, pg. 13.

Sneddon, pg. 31.